A lift is going upwards with an acceleration of ` 4.9 m//s^(2) ` What will be the apparent weight of a 60 kg person sitting in the lift ? What when the lift acquires a unifrom velocity of ` 4.9 m//s ` ? If rope of the lift is broken , then ? Take ` g = 9.8 m//s^(2)` .

Explanation

Here , ` a = 4.9 m//s^(2) , m = 60 kg R=? `
As lift is accelerating upwards therefore, apparent weight ` R = m (g + a ) `
` = 60 (9.8 + 4.9) N `
` R = (60xx14.7)/(9.8) kg f = 90 kg f `
(ii) When lift acquires a unifrom velocity , a = 0
` :. R = m (g + a ) = 60 xx 9.8 N = 60 kg f `
(iii) If rope of lift is broken , a = g downwards
` :. R' = m (g - a ) = m (g - g ) = Zero ` .